Identify the ionizable groups in histidine

Histidine contains three important ionizable sites:

• Carboxyl group: $-COOH \rightleftharpoons -COO^- + H^+$
• Amino group: $-NH_3^+ \rightleftharpoons -NH_2 + H^+$
• Imidazole side chain: can also be protonated

Recall their approximate $pK_a$ values

For histidine, typical values are:

• $pK_a(\ce{-COOH}) \approx 1.8$
• $pK_a(\ce{imidazole}) \approx 6.0$
• $pK_a(\ce{-NH3+}) \approx 9.2$

Analyze the situation at $\mathbf{pH = 2}$

Rule:

• If $\text{pH} < pK_a$, the group is predominantly protonated.
• If $\text{pH} > pK_a$, the group is predominantly deprotonated.

Now check each group:

• Amino group: since $2 \ll 9.2$, it remains protonated as $\ce{-NH3+}$.
• Imidazole side chain: since $2 \ll 6.0$, it is also protonated, carrying a positive charge.
• Carboxyl group: since $2$ is very close to $1.8$, but in a strongly acidic medium it is taken predominantly as protonated $\ce{-COOH}$.

Hence the overall structure at pH 2

Histidine in strongly acidic solution will have:

• $\ce{-NH3+}$ on the amino group
• protonated imidazole ring ($+1$ charge on side chain)
• $\ce{-COOH}$ on the carboxyl group

So the species is a dicationic form overall.

Match with the correct option

Therefore, the correct option must be the one showing:

$\ce{H3N+ - CH(CH2-imidazolium) - COOH}$

with the imidazole nitrogen protonated.

This corresponds to Option B.

Comparison with stored answer

Stored correct answer = B

Our derived answer = B

So they agree.